# Probability of rolling 3 mutually exclusive dice

Suppose I have 3 dice. Each has some mechanism that can prevent other dice being in the same number as itself when they are rolled together. Now I roll the 3 dice at the same time. Then what is the probability of having a 2 appear?

Is the answer 1/6 + 1/6 + 1/6 = 1/2 since the event are mutually exclusive and each has a probability of 1/6? But 1/2 seems to be too high?

• What do you mean "at least"??? According to your description, there cannot be more than a single occurrence of each value. – barak manos Sep 1 '14 at 12:42
• thank you, it should be there is a 2 appear – Allitee Sep 1 '14 at 12:49
• This question has generated a kilometre of LaTeX! But the answer is obvious: If exactly three different numbers are rolled, the probability of a specific number belonging to those three is $\frac36$. – TonyK Sep 1 '14 at 12:51
• @TonyK: Nice! That's actually a cleaner explanation than the "traditional" combinatorics type of answers (such as my own answer in this case). – barak manos Sep 1 '14 at 13:02
• ADD: If the 3 dice can only prevent 2 from from appearing in other dice if itself is 2 but not other number. Should the probability change? Is the event of 2 appearing in each of three dice still mutually exclusive. – Allitee Sep 1 '14 at 13:59

Because a dice prevents the same number from appearing on another dies means that you can roll only those combinations which contain distinct numbers. The number of such combinations is $^6\mathrm{C}_3 \times 3!$

Number of such combinations with $2$ in them are $^5\mathrm{C}_2 \times 3!$

So the probability that $2$ appears on one of the dies is $\dfrac{^5\mathrm{C}_2 \times 3!}{^6\mathrm{C}_3 \times 3!} = \dfrac{1}{2}$

Yes the answer is $\frac{1}{2}$

Think of it as selecting 3 items in the set $\{1,2,3,4,5,6\}$ which can be done in $6\choose3$$=20 ways. Convince yourself that this is the sample space (and not 6^3) Now, at least one 2, means that there is ONLY one 2 and 2 always appears (since these magical die makes sure 2 doesnt appear in the other two) That would leave us with choosing 2 out of \{1,3,4,5,6\} which is 5\choose2$$ = 10$

The probability is $\frac{10}{20} = \frac{1}{2}$

$1/2$ is most definitely very high, but you are thinking about the sample space $6^3$. Out of 20 possibilities 2 appears 1/2 the time, which makes sense because, each number either appears or does not (because of the mutual-exclusiveness) and if you calculate for other numbers (1,3,...) you'll notice that the answer is 1/2 as well.

Note: Im assuming the order of the die don't matter. If they do, multiply and divide by $3!$

The total number of combinations is $\dbinom{6}{3}\cdot3!=120$

The number of combinations with 2 in them is $\dbinom{5}{2}\cdot3!=60$

So the probability of getting a combination with 2 in it is $\dfrac{60}{120}=\dfrac{1}{2}$